Alternative English Year - 2017 Full Marks : 100
Class 12 Ncert Solution Maths|| 3d Geometry
Class 12 Ncert Solution Maths|| 3d Geometry
Exercise 11.2
1. Show that the three lines with direction cosines
\(\frac{12}{13}, \ \frac{-3}{13},\ \frac{-4}{13};\ \ \frac{4}{13},\ \frac{12}{13},\ \frac{3}{13};\ \ \frac{3}{13},\ \frac{-4}{13},\ \frac{12}{13}\) are mutually perpendicular.
Sol: Two lines are perpendicular if the sum of the product of their direction cosines is equal to zero.
If \(l_1,\ m_1,\ n_1\) and \(l_2,\ m_2,\ n_2\) be the direction cosines of two lines then they are perpendicular if \(l_1l_2 + m_1m_2 + n_1n_2=0\)
\(\therefore\)\(l_1l_2 + m_1m_2 + n_1n_2\)
=\(\frac{12}{13}.\frac{4}{13} + \frac{-3}{13}.\frac{12}{13} +\frac{-4}{13}.\frac{3}{13}\)
= \(\frac{48}{169} - \frac{36}{169} -\frac{12}{169}\)
=0
Thus the lines having direction cosines \(\frac{12}{13}, \ \frac{-3}{13},\ \frac{-4}{13};\ \ \frac{4}{13},\ \frac{12}{13},\ \frac{3}{13}\) are perpendicular.
\(l_1l_2 + m_1m_2 + n_1n_2\)
=\(\frac{4}{13}.\frac{3}{13} + \frac{12}{13}.\frac{-4}{13} +\frac{3}{13}.\frac{12}{13}\)
= \(\frac{12}{169} - \frac{48}{169} +\frac{36}{169}\)
=0
Thus the lines having direction cosines \(\frac{4}{13},\ \frac{12}{13},\ \frac{3}{13};\ \ \frac{3}{13},\ \frac{-4}{13},\ \frac{12}{13}\) are mutually perpendicular.
\(l_1l_2 + m_1m_2 + n_1n_2\)
=\(\frac{3}{13}.\frac{12}{13} + \frac{-4}{13}.\frac{-3}{13} +\frac{-4}{13}.\frac{12}{13}\)
= \(\frac{36}{169} + \frac{12}{169} -\frac{48}{169}\)
=0
The two lines having direction cosines \(\frac{3}{13},\ \frac{-4}{13},\ \frac{12}{13};\ \frac{12}{13},\ \frac{-3}{13},\ \frac{-4}{13}\) are perpendicular.
Thus all the three lines are mutually perpendicular.
2. Show that the line through the points(1,-1,2), (3,4,-2) is perpendicular to the line through the point(0,3,2) and (3,5,6).
Sol: Direction ratio of the line passing through the points (1,-1,2), (3,4,-2) are
(3-1),(4+1),(-2-2)
i.e. 2, 5, -4
Direction ratio of the line passing through the points (0, 3, 2), (3, 5, 6) are
(3-0),(5-3),(6-2)
i.e. 3, 2, 4.
\(\therefore\)\(l_1l_2 + m_1m_2 + n_1n_2\)
=2.3 +5.2+(-4)4
=6 + 10 -16
=0
Hence , the line through the points(1,-1,2), (3,4,-2) is perpendicular to the line through the point(0,3,2) and (1, 2, 5).
3. Show that the line through the points (4, 7, 8), (2, 3, 4) is parallel to the line through the points (-1, -2, 1), (3, 5, 6)
Sol: Direction ratio of the line through (4, 7, 8), (2, 3, 4) is
(2-4), (3-7), (4-8)
i.e. (-2, -4, -4)
Direction ratio of the line through (-1, -2, 1), (1, 2, 5) is
(1+1), (2+2), (5-1)
i.e. 2, 4, 4
The direction ratios are proportional so it is clear that the lines are parallel.
Hence, the line through the points (4, 7, 8), (2, 3, 4) is parallel to the line through the points (-1, -2, 1), (3, 5, 6)
4. Find the equation of the line which passes through the point (1, 2, 3) and is parallel to the vector \(3\hat i + 2\hat j - 2\hat k\)
Sol: The position vector of the point (1, 2, 3) is \(\hat i + 2\hat j + 3\hat k\) .Let it be denoted by \(\vec a\). Thus, \(\vec a\) = \(\hat i + 2\hat j + 3\hat k\)
Let the other vector to which the line is parallel be denoted by \(\vec b\).
Thus, \(\vec b\) = \(3\hat i + 2\hat j - 2\hat k\)
The line which passes through the given point and parallel to \(\vec b\) is given by
\begin{align}
\vec r &= \vec a + \lambda \vec b\\
\vec r &= (\hat i +2\hat j + 3\hat k) + \lambda (3\hat i +2\hat j - 2\hat k)\\
\end{align}
this is the required equation of the line
6. Find the cartesian equation of the line which passes through the point (-2,4,-5) and parallel to the line given by \(\frac{x+3}{3}=\frac{y-4}{5}=\frac{z+8}{6}\)
Sol: The line passing through a point \(x_1,y_1,z_1\) and having direction ratios a. b and c is given by
\(\frac{x-x_1}{a}=\frac{y-y_1}{b}=\frac{z-z_1}{c}\)
The required line is parallel to the line \(\frac{x+3}{3}=\frac{y-4}{5}=\frac{z+8}{6}\) (this line has direction ratios 3, 5 and 6)
Thus, the direction ratios of the required line is proportional to 3, 5 and 6.
Hence, the direction ratio of the required line is 3k, 5k and 6k.
Therefore , cartesian equation of the line which passes through the point (-2,4,-5) and parallel to the line given by \(\frac{x+3}{3}=\frac{y-4}{5}=\frac{z+8}{6}\) is
\vec r &= \vec a + \lambda \vec b\\
\vec r &= (\hat i +2\hat j + 3\hat k) + \lambda (3\hat i +2\hat j - 2\hat k)\\
\end{align}
this is the required equation of the line
6. Find the cartesian equation of the line which passes through the point (-2,4,-5) and parallel to the line given by \(\frac{x+3}{3}=\frac{y-4}{5}=\frac{z+8}{6}\)
Sol: The line passing through a point \(x_1,y_1,z_1\) and having direction ratios a. b and c is given by
\(\frac{x-x_1}{a}=\frac{y-y_1}{b}=\frac{z-z_1}{c}\)
The required line is parallel to the line \(\frac{x+3}{3}=\frac{y-4}{5}=\frac{z+8}{6}\) (this line has direction ratios 3, 5 and 6)
Thus, the direction ratios of the required line is proportional to 3, 5 and 6.
Hence, the direction ratio of the required line is 3k, 5k and 6k.
Therefore , cartesian equation of the line which passes through the point (-2,4,-5) and parallel to the line given by \(\frac{x+3}{3}=\frac{y-4}{5}=\frac{z+8}{6}\) is
\(\frac{x-(-2)}{3k}=\frac{y-4}{5k}=\frac{z-(-5)}{6k}\)
\(\implies\)\(\frac{x+2}{3k}=\frac{y-4}{5k}=\frac{z+5}{6k}\)
\(\implies\)\(\frac{x+2}{3}=\frac{y-4}{5}=\frac{z+5}{6}\)
7. The cartesian equation of aline is \(\frac{x-5}{3}=\frac{y+4}{7}={z-6}{2}\)
Sol: It is clear that the given line passes through the point (5, -4, 6).
\(\therefore\) Position vector of the point (5, -4, 6) is \(\vec a =5\hat i -4\hat j+6\hat k\)
From the equation of the line itcan be seen that its direction ratios are 3, 7 and 2.
Thus the line parallel to the given line is \(\vec b =3\hat i + 7\hat j+ 2\hat k\)
Hence the equation of the line in vector form is
\(\vec r= 5\hat i -4\hat j+6\hat k + \lambda (3\hat i + 7\hat j+ 2\hat k) \)
8. Find the vector and the cartesian equations of the line that passes through the origin and (5, -2, 3).
Sol: The required line passes through (0,0,0) and (5,-2,3). So the direction ratios of this line are(5-0),(-2-0),(3-0) i.e 5, -2, 3.
Hence the equation of the line passing through (0,0,0) and having direction ratio 5, -2, 3 is given by
\(\frac{x-0}{5}=\frac{y-0}{-2}=\frac{z-0}{3}\)
\(\frac{x}{5}=\frac{y}{-2}=\frac{z}{3}\)
The position vector for the point (0,0,0) is \(\vec a\)=0 and the position vector for the point (5,-2, 3) is \(\vec b=5\hat i -2\hat j +3\hat k\)
The equation of a line in vector form is given by
\(\vec r = \vec a +\lambda (\vec b\)
\(\vec r = 0 +\lambda (5\hat i -2\hat j +3\hat k\)
\(\vec r= \lambda (5\hat i -2\hat j +3\hat k)\)
9. Find the vector and cartesian equations of the line that passes through the points (3, -2, -5). (3, -2, 6)
Sol: The required line passes through (3, -2, -5) and (3, -2, 6). So the direction ratios of this line are(3-3),(-2-(-2)),(6-(-5)) i.e 0, 0, 11.
Hence the equation of the line passing through (3, -2, -5) and having direction ratio 0, 0, 11 is given by
\(\frac{x-3}{0}=\frac{y-(-2)}{0}=\frac{z-(-5)}{11}\)
\(\frac{x-3}{0}=\frac{y+2}{0}=\frac{z+5}{11}\)
The position vector for the point (3, -2, -5) is \(\vec a= 3\hat i-2\hat j-5\hat k\) and the position vector for the point (0, 0 , 11) is \(\vec b=11\hat k\)
The equation of a line in vector form is given by
\(\vec r = \vec a +\lambda \vec b\)
\(\vec r = 3\hat i-2\hat j-5\hat k+\lambda (11\hat k)\)
10. Find the angle between the following pair of lines:
(i) \(\vec r = 2\hat i -5\hat j +\hat k +\lambda(3\hat i +2\hat j +6\hat k)\) and
\(\vec r =7\hat i -6\hat k +\mu(\hat i + 2\hat j +2\hat k)\)
(ii) \(\vec r =3\hat i +\hat j -2\hat k + \lambda (\hat i -\hat j-2 \hat k)\) and
\(\vec r =2\hat i -\hat j -56\hat k +\mu(3\hat i -5\hat j -4\hat k)\)
Sol: Angle between two lines is given by
\(\cos \theta =\left|\frac{\vec b_1.\vec b_2}{|\vec b_1|.|\vec b_2|}\right|\)
(i)
Given that \(\vec b_1= 3\hat i +2\hat j +6\hat k\) and \(\vec b_2=\hat i + 2\hat j +2\hat k\)
\(\therefore\) \(|\vec b_1|\) = \(\sqrt{3^2 + 2^2 +6^2}=\sqrt{49}=7\)
and \(|\vec b_2|\) = \(\sqrt{1^2 + 2^2 +2^2}=\sqrt{9}=3\)
Hence, \(\cos \theta =\left|\frac{(3\hat i +2\hat j +6\hat k).(\hat i + 2\hat j +2\hat k)}{7.3}\right|\)
= \(\left|\frac{3+4+12}{21}\right|=\frac{19}{21}\)
Thus, \(\theta=\cos^{-1}\left(\frac{19}{21}\right)\)
(ii) Given that \(\vec b_1= \hat i -\hat j -2\hat k\) and \(\vec b_2=3\hat i - 5\hat j -4\hat k\)
\(\therefore\) \(|\vec b_1|\) = \(\sqrt{1^2 + 1^2 +2^2}=\sqrt{6}\)
and \(|\vec b_2|\) = \(\sqrt{3^2 + 5^2 + 4^2}=\sqrt{50}=5\sqrt{2}\)
Hence, \(\cos \theta =\left|\frac{(\hat i -\hat j -2\hat k).(3\hat i - 5\hat j -4\hat k)}{\sqrt{6}.5\sqrt{2}}\right|\)
= \(\left|\frac{3+5+8}{10\sqrt{3}}\right|=\frac{16}{10\sqrt{3}}=\frac{8}{5\sqrt{3}}\)
Thus, \(\theta=\cos^{-1}\left(\frac{8}{\sqrt{3}}\right)\)
11. Find the angle between the following pair of lines:
(i) \(\frac{x-2}{2}=\frac{y-1}{5}=\frac{z+3}{-3}\) and \(\frac{x+2}{-1}=\frac{y-4}{8}=\frac{z-5}{4}\)
(ii) \(\frac{x}{2}=\frac{y}{2}=\frac{z}{1}\) and \(\frac{x-5}{4}=\frac{y-2}{1}=\frac{z-3}{8}\)
Sol: The angle between two lines having direction ratios \(a_1,b_1,c_1 \ and\ a_2\), \(b_2 \ and\ c_2\) is given by \(\cos \theta = \left|\frac{a_1a_2 +b_1b_2 +c_1c_2}{\sqrt{a_1^2 + b_1^2 + c_1^2}\sqrt{a_2^2 +b_2^2 + c_2^2}}\right|\)
(i) The angle between the given two lines is given by
\(\cos \theta = \left|\frac{2(-1) +5.8 + 4(-3)}{\sqrt{2^2 + 5^2 + (-3)^2}\sqrt{(-1)^2 +8^2 + 4^2}}\right|\)
= \( \left|\frac{-2 +40 -12}{\sqrt{4 + 25 + 9}\sqrt{1 +64 + 16}}\right|\)
=\(\frac{26}{\sqrt{38}\sqrt{81}}\)
\(\theta =\cos^{-1}\left(\frac{26}{9\sqrt{38}}\right)\)
(ii) The angle between the given two lines is given by
\(\cos \theta = \left|\frac{2.4 +2.1 + 1.8}{\sqrt{2^2 + 2^2 + 1^2}\sqrt{4^2 +1^2 + 8^2}}\right|\)
= \( \left|\frac{8 +2 +8}{\sqrt{4 + 4 + 1}\sqrt{16 + 2 + 64}}\right|\)
=\(\frac{18}{\sqrt{9}\sqrt{81}}\)
=\(\frac{18}{3\times9}\)
\(\theta =\cos^{-1}\left(\frac{2}{3}\right)\)
12. Find the values of p so that the lines \(\frac{1-x}{3}=\frac{7y-14}{2p}=\frac{z-3}{2}\) and
\(\frac{7-7x}{3p}=\frac{y-5}{1}=\frac{6-z}{5}\) are at right angles.
Sol: The given two lines will be at right angles when the sum of the product of the direction ratios is equal to zero. i.e \(a_1a_2 + b_1b_2 + c_1c_2 =0\)
\(\therefore\) \((-3)\frac{-3p}{7} + \frac{2p}{7}1 + 2\times (-5) =0\)
\(\implies \frac{9p}{7} +\frac{2p}{7} -10=0\)
\(\implies 11p=10\)
\(\implies p=\frac{10}{11}\)
13. Show that the lines \(\frac{x-5}{7}=\frac{y+2}{-5}=\frac{z}{1}\) and \(\frac{x}{1}=\frac{y}{2}=\frac{z}{3}\) are perpendicular to each other.
Sol: The given two lines will be perpendiculkar to each other when the sum of the product of the direction ratios is equal to zero. i.e \(a_1a_2 + b_1b_2 + c_1c_2 =0\)
\(\therefore\) L.H.S = \(7\times1+(-5)\times 2+ 1\times3 \)
=7-10+3
=0 =R.H.S
Since the sum is equal to zero so the lines are perpendicular.
Ncert Solutions for other subjects.
General English
A Thing of Beauty
The Roadside Stand
The Last Lesson
Going Places
The memoirs of the Chota Sahib
Question paper for class 12 AHSEC
Chemistry
Physics
Maths
Alternative English 2016
Alternative English 2017
Class 12 Ncert Solution Maths|| 3d Geometry
Previous Exercise
Next Exercise
Ncert Solutions for other subjects.
A Thing of Beauty
The Roadside Stand
The Last Lesson
Going Places
The memoirs of the Chota Sahib
Question paper for class 12 AHSEC
Chemistry
Physics
Maths
Alternative English 2016
Alternative English 2017
Comments
Post a Comment